SnakeTongue
Graduate Poster
- Joined
- Aug 16, 2010
- Messages
- 1,084
So, let's revisit the different models and sum up:
1. You accept Provan's experiment and have conceded you were flat out lying by saying that "Charles D Provan’s experiment demonstrated an average body of 0.07158 cubic meters" because it quite clearly demonstrated 18.2 bodies/cubic metres and not 13.97 bodies/cubic metre
2. You have yet to offer any convincing argument as to why one cannot take average weights and apply them to the hypothetical Bay model.
3. You shot yourself in the foot by fuss-making over Roberto using a lower average weight for 'Vitruvian men' and now repeat my calculation that the density would be 0.7 tons/cubic metre, not 0.66 tons/cubic metre
4. That density, when divided by weight only, produces more than 20 bodies/cubic metre.
5. Roberto's division of his lower density of 0.66 tons/cubic metre by 34kg, produces 19.51 bodies/cubic metre.
6. The difference between 19.51 bodies/cubic metre, as arrived at by Roberto's use of Bay's hypothetical model and Roberto's own hypothetical weight estimates, and the result of Provan's experiment, 18.2 bodies/cubic metre, is insignificant. Provan's experiment thus confirms the order of magnitude of the hypothetical calculation based on Bay and dividing by weight only.
7. The hypothesised average weight of 34kg is not simply based on the extremely crude calculation of two adults plus a child, but was also approximated Gerstein during his 1945 interrogations when he estimated an average weight of 35kg. The average weight of Provan's experimental group was 33.25 kg, confirming Gerstein and supporting the hypothetical average weight.
8. Your hocus-pocus with the average weights ignores the fact that the hypothetical 2 adults + 1 child was a crude approximation; there are many other ways of combining different age groups across both genders to arrive at an extraordinarily low average weight, as demonstrated by Provan's experiment. In essence, your dishonesty is to treat placeholder approximations as fixed elements, rather than recognise that they were actually placeholders for the purpose of doing a simplified calculation.
9. Therefore, the number of bodies that could have fitted into the available grave space was far closer to the total number of victims than you alleged.
10. The difference, acknowledged ever since 2006 by Roberto when he started looking at this issue, is explained by the effects of decomposition and grave-settling, along with the fact that Belzec shut down at the end of December 1942 because the graves were overflowing.
11. Other things you ignore: it is possible that other graves were not located by Kola, as Bay has argued, but we don't know. It is certain based on documentary evidence that several thousands of the 434,000 recorded victims never reached the camp at all because they died en route, were offloaded from trains or jumped from trains, and were buried elsewhere. This number will be at least 1% of the victims.
12. Your English sucks.
1) The two different results come from two different body distribution ratios. I calculated Roberto Muehlenkamp’s 2:1 ratio and Charles D Provan’s 3:4:1 ratio. Both calculations use a single formula based on Charles D Provan’s experiment and both results are accurate. Your comparison using an inexistent scalar/measurement is clueless. The accusation “flat out lying” is just foolish and you did not provide any arithmetical evidence to support it.
2) The Charles A Bay study only regards the proportions of a male model to estimate the capacity of a hypothetical mass grave. To apply mass to his experiment is irrelevant for the reason that the model volume estimation does not rely on mass variation. The imaginary model was only produced to fit a hypothetical mass grave. The study does not offer reliable data which could be used to safely define the approximate mass of different hypothetical models.
3) That does not validate Roberto Muehlenkamp’s method. It only shows that an imaginary burial pit filled with hypothetical 66Kg male models has similar density as verified in a garbage dump landfill. It does not support Roberto Muehlenkamp’s calculations; neither establishes the body volume of another model.
4) When density is divided by mass it results in reciprocal cubic meter, not in “bodies/cubic metre”:
706Kg/m^3 / 34Kg = 20.76m^-3 (reciprocal cubic meters)
5) Let’s test your recalcitrant application of Roberto Muehlenkamp’s method:
36 bodies (1.20m 35Kg) / 2.0m^3 = 18 bodies per cubic meter
09 bodies (1.20m 15Kg) / 0.5m^3 = 18 bodies per cubic meter
The above situations results in exactly the same number of bodies per cubic meter. However, each situation is represented by bodies with different mass. Thus the hypothetical density for each situation would be:
18 bodies per cubic meter * 35Kg = 630Kg/m^3
18 bodies per cubic meter * 15Kg = 320Kg/m^3
Assuming Roberto Muehlenkamp’s method, one would expect using any given situation to guess the bodies per cubic meter of another situation. In other words, if 18 bodies with 35Kg fit one cubic meter, then how many bodies with 15Kg would fit one cubic meter?
630Kg/m^3 / 15Kg = 42 bodies per cubic meter
So, why the result is 42 bodies per cubic meter? It should be 18 bodies per cubic meter, since one situation demonstrated that:
09 bodies (1.20m 15Kg) / 0.5m^3 = 18 bodies per cubic meter
Let’s use the last situation:
320Kg /m^3 / 35Kg = 9.1 bodies per cubic meter
So, why the result is 9.1 bodies per cubic meter? It should be 18 bodies per cubic meter, since one situation demonstrated that:
36 bodies (1.20m 35Kg) / 2.0m^3 = 18 bodies per cubic meter
Which situation above is true and why?
6) Roberto Muehlenkamp’s calculation does not have any “order of magnitude”.
7) It is just based on a simple calculation of sample mean. Roberto Muehlenkamp underestimated the average mass of a given group of people. The anthropological data is valid and offers consistent data. However, the statistical data from a single witness is not compelling evidence to support the hypothesis that Polish ghetto Jews were extremely underweight. The experiment of Charles D Provan cannot be used to support such hypothesis because the participants were not particularly underweight.
Pictures from Polish people living in ghettos:
http://historyimages.blogspot.com.br/2012/02/poland-under-german-occupation-and.html
8) I did not “ignore” any essential value provided by Roberto Muehlenkamp:
(...) Assuming that the average weight of adult Jews in Polish ghettos at the time was in between the upper and the lower value of what the BMI table considers underweight, it would be (38+48) ÷ 2 = 43 kg. (...) This relation would mean a weight of 43 ÷ 2.76 = 15.6 kg for ill-fed or starving children in Polish ghettos. (...) Rounding up the latter value, a group of two adults and one child 14 years and younger from a Jewish ghetto in Poland would thus weigh (43+43+16)/3 = 34 kg on average, (...)
SnakeTongue said:The formula can be applied to the Holocaust Controversies estimations of 3 adults with a total mass (a) of 129Kg (3*43Kg), 4 children with a total mass (b) of 64Kg (4*16Kg) and 1 toddler with a total mass (c) of 16Kg (1*16Kg):
{x = a/(a+b+c)*0.44/3, y = b/(a+b+c)*0.44/4, z = c/(a+b+c)*0.44/1}
{x = 129/(129+64+16)*0.44/3, y = 64/(129+64+16)*0.44/4, z = 16/(129+64+16)*0.44/1}
Using the Holocaust Controversies distribution of 2 adults and 1 child the total volume of all bodies is:
V = 0.0905263 + 0.0905263 + 0.0336842
V = 0.2147368m^3
The average body volume of 2 adults and 1 child is 0.07158 cubic meters. Thus a 21,310 cubic meters burial pit would hold up to 297,713 bodies of adults and children with an average weight of 34 kilograms.
9) Not so close...
10) Human bodies have a low decomposition rate after buried. The mass graves were filled and sealed approximately every 8 days. Under such conditions the bodies would not decompose fast enough to free considerable space.
11) I do not ignore such possibilities. If 434,508 deportees were killed, definitely not all bodies were buried in Belzec mass graves as I have demonstrated.
12) Whatever... Native English speakers cannot even properly spell the word “cão” (dog) after a long time of practising. Try it:
“CãoZilla”
(...) But to be honest, I took more than a year to be able to say it like a Brazilian. (...) However, ã is usually followed by an o (ão). Together, these letters make an ah-ooh sound. But say it fast, and you say Ow! like you’ve hurt yourself. Brazilians say the ã like the English ow, only with the nasal sound you just practiced.
Karen Keller, Portugues Phrase for Dummies, pag. 15-16
Karen Keller, Portugues Phrase for Dummies, pag. 15-16
Go **** yourself. There is absolutely nothing wrong with estimating the average number of bodies per cubic metre. Your babble about 'scalars' and so forth are just gibberish.

Reciprocal length or inverse length is a measurement used in several branches of science and mathematics.
http://en.wikipedia.org/wiki/Reciprocal_length
http://en.wikipedia.org/wiki/Reciprocal_length